What is the acceleration of a buoyant alcohol-filled shell in water?

[AG1189]

Homework Statement

-A thin spherical shell of mass 0.300 kg and diameter 0.250 m is filled with alcohol ( = 806 kg/m3). It is then released from rest on the bottom of a pool of water. Find the acceleration of the alcohol filled shell as it rises toward the surface of the water.

Homework Equations

FB-(M1+M2)g=(M1+M2)a

The Attempt at a Solution

M1=0.3kg, r=0.125 m, V= (4*pi*r^3) /3,
M2 = pV = (806*V), FB = pVg = (1000)(V)(9.8)
. After finding FB, I plugged it back into the equation and got 2.211 m/s^2. This answer was wrong, could someone please tell me if I am using the wrong equation, or if I just screwed something up.

 

[gabee]

Homework Statement

-A thin spherical shell of mass 0.300 kg and diameter 0.250 m is filled with alcohol ( = 806 kg/m3). It is then released from rest on the bottom of a pool of water. Find the acceleration of the alcohol filled shell as it rises toward the surface of the water.

Homework Equations

FB-(M1+M2)g=(M1+M2)a

The Attempt at a Solution

M1=0.3kg, r=0.125 m, V= (4*pi*r^3) /3,
M2 = pV = (806*V), FB = pVg = (1000)(V)(9.8)
. After finding FB, I plugged it back into the equation and got 2.211 m/s^2. This answer was wrong, could someone please tell me if I am using the wrong equation, or if I just screwed something up.

It looks like you might have just done something wrong in your calculations...I set it up like this:

$$\sum{F} = \rho_{water} V g - \rho_{alcohol} V g - mg = (m + \rho_{alcohol} V) a$$

$$a = \frac{g(\frac{4}{3} \pi r^3 (\rho_{water} - \rho_{alcohol}) - m)}{m + \rho_{alcohol} \frac{4}{3} \pi r^3}$$

and I got a number a little lower than yours. Maybe it was rounding? Generally I like to get everything in one equation and then plug it all in at once to reduce the chance of that sort of mistake.

 

[AG1189]

:)

it worked out, thank you very much. I got 1.829711 m/s^2, I think that's what you got.